13 Sided Polygon of Charges

  • #1
Imagine you have a regular 13 sided polygon with charges distributed on every corner of the polygon.

What would a test charge experience in the center?

The answer to that was a 0 net force (which makes some intuitive sense to me due to the symmetry of the polygon). I understand that if we were to place a force in the

However what would happen if one were to remove one of the charges?

  • #2
Just draw the 12 arrows and you´ll see the result.
  • #3
Just draw the 12 arrows and you´ll see the result.
That seems a rather tedious and time-consuming way to go about this. It is much easier to take the previous result of zero, then subtract the force that was due to the removed charge.
  • #4
Alright that seems like a reasonable way to go about this.. I guess I just want to double check too that describing the symmetry of the polygon is an adequate way to solve this problem, and if there was a more elegant way to answer the problem.
  • #5
Just draw the 12 arrows and you´ll see the result.
After thinking more, actually this is not difficult. The 12 arrows form 12 sides of a 13-sided regular polygon; the missing side gives the resulting force. Nice way to look at it!
  • #6
and a nicee idea!
imagine we have an N gon on xy page.we want to find the E vector at some z(out the page over the center of polygon) but not actually OVER the center.say we move away from the axis z(which starts at O which is the center of polygon) a tiny length delta .it means in cylindrical coordinates we are at z(z) and delta(r) and some phi(which won't matter).to the first nonzero order of delta we want the E vector.
the idea:finding E on the z axis is fairly easy.(because of symmetry).the trick is to take a differential cylinder near the point on the axis and write divergence>>everything going in comes out.things which go in is Ez from below and Ez+dz comes out .also from the sides of cylinder E r comes out.(we want E r)equating the flux simply gives the desired E.nicee!

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